math20c Meesue Yoo fall 09 quiz3

# math20c Meesue Yoo fall 09 quiz3 - 2 t = 4 So s = Z t 4 du...

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Solution of Quiz 3 for MATH 20C, 2009 1 . Describe the domain and the range of the function f ( x,y,z ) = p 9 - x 2 - y 2 - z 2 Ans. Domain of f = { ( x,y,z ) | 9 - x 2 - y 2 - z 2 0 } = { ( x,y,z ) | x 2 + y 2 + z 2 9 } which is the interior and the boundary of a sphere centered at the origin with radius 3. For the range of f , if we let w = x 2 + y 2 + z 2 , by the domain of f , 0 w 9, so 0 9 - w 3. Hence, Range of f = { w | 0 w 3 } . 2 . Find an arc length parametrization of the circle in the plane z = 9 with radius 4 and center (1 , 4 , 9). Note that the arc length function is deﬁned by s ( t ) = Z t 0 || r 0 ( u ) || du. Ans. A parametrization of a circle with radius 4 centered at (1 , 4 , 9) on z = 9 is r ( t ) = < 4 cos t + 1 , 4 sin t + 4 , 9 > . r 0 ( t ) = < - 4 sin t, 4 cos t, 0 > || r 0 ( t ) || = p 16 sin 2 t + 16 cos
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Unformatted text preview: 2 t = 4 . So, s = Z t 4 du = 4 t ⇒ t = s 4 . Hence, the arc length parametrization is r 1 ( s ) = r ( s/ 4) = D 4 cos ± s 4 ² + 1 , 4 sin ± s 4 ² + 4 , 9 E . 3 . Find r ( t ) and v ( t ) given a ( t ) and the initial velocity and position a ( t ) = t k , v (0) = i , r (0) = j . Ans. v ( t ) = Z t a ( s ) ds + v (0) = ³Z t sds ´ k + i = i + 1 2 t 2 k = µ 1 , , 1 2 t 2 ¶ . r ( t ) = Z t v ( s ) ds + r (0) = ·³Z t 1 ds ´ i + ³Z t 1 2 s 2 ds ´ k ¸ + j = t i + j + 1 6 t 3 k = µ t, 1 , 1 6 t 3 ¶ ....
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## This note was uploaded on 01/26/2010 for the course MATH Math 20C taught by Professor Lunasin during the Fall '08 term at UCSD.

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